is this a way to work this out? Solve S = pi r sqrt{r^2 + h^2} for h

[allegansveritatem]

Here is the problem:

Here is my solution:


Here is what my calculator says:


If calculator is right here, where did it get that r^4?

 

[Dr.Peterson]

Try carrying out the addition of fractions under the radical, using a common denominator. You'll see the r^4 appear.

But note that R and r, H and h are different variables. Don't get in the habit of changing the case of variables when you write; you will eventually get caught with an equation involving both cases, and will stumble!

 

[allegansveritatem]

Try carrying out the addition of fractions under the radical, using a common denominator. You'll see the r^4 appear.

But note that R and r, H and h are different variables. Don't get in the habit of changing the case of variables when you write; you will eventually get caught with an equation involving both cases, and will stumble!

Thanks. I will try it tomorrow. I can see right now that operating as you suggest is going to change the game significantly. Yes, I will be more careful of my capitals.

 

[allegansveritatem]

So, I went and redid this and and got a r^4 but I don't know that I improved my condition any with it. Here is the result I have so far. Does it look close?

 

[allegansveritatem]

Wow! I just saw that all that stuff on the RS is able to be removed from under the radical sign with only the r in the numerator left with an exponent (greater than 1)and we are done! Happy me!

 

[Steven G]

Unfortunately what you said is not true. Can we please see your work so we can figure out what happened? Remember that sqrt (a - b)\(\displaystyle \neq\) sqrt(a) - sqrt (b)

 

[Dr.Peterson]

Wow! I just saw that all that stuff on the RS is able to be removed from under the radical sign with only the r in the numerator left with an exponent (greater than 1)and we are done! Happy me!

since you previously showed the correct answer, which includes a radical, I am assuming that you didn't really mean that everything comes out from the radical, but just that the denominator does, so you got what the calculator said. Am I right?

 

[allegansveritatem]

Unfortunately what you said is not true. Can we please see your work so we can figure out what happened? Remember that sqrt (a - b)\(\displaystyle \neq\) sqrt(a) - sqrt (b)

I will get back to this. I am not sure what you mean.

 

[allegansveritatem]

since you previously showed the correct answer, which includes a radical, I am assuming that you didn't really mean that everything comes out from the radical, but just that the denominator does, so you got what the calculator said. Am I right?

I will get back to this later tonight. I have to look at this again and get back into it to know what to answer.

 

[allegansveritatem]

Here is what I end up with. I don't know if it is the same as the calculator or not....seems the calculator is offering several versions.



I have no great faith in this...but seems defensible so far, no?

 

[Dr.Peterson]

Here is what I end up with. I don't know if it is the same as the calculator or not....seems the calculator is offering several versions.

View attachment 11650

I have no great faith in this...but seems defensible so far, no?

The last step is what Jomo said is wrong. You can't take the square root of each term of a sum; [MATH]\sqrt{a^2 - b^2} \ne a - b[/MATH], so [MATH]\sqrt{S^2 - \pi^2r^4} \ne S - \pi r^2[/MATH]. The equation before that is what your calculator showed, and is correct. (The inequalities it showed after the solution are just conditions for its being evaluated, and are oddly written; you can ignore those.)